How do you integrate $e^{tan x} \cdot {sec}^{2} (x) d x$ $e^(tanx) * sec^2(x) dx$ ?

Question

36338 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-12T04:33:25

$\int e^{tan x} {sec}^{2} (x) d x = e^{tan x} + C$ $int e^(tan x) sec^(2)(x) dx = e^(tan x) + C$

We can use a simple u-substitution to evaluate this integral.

Given $\int e^{tan x} {sec}^{2} (x) d x$ $int e^(tan x) sec^(2)(x) dx$

Note that ${sec}^{2} (x)$ $sec^2(x)$ is the derivative of $tan x$ $tan x$ , thus we can use

$u = tan x \to d u = {sec}^{2} (x) d x$ $u = tan x -> du = sec^(2)(x) dx$

Since

$\int e^{u} d u = e^{u} + C$ $int e^(u) du = e^(u) + C$

We only have to substitute $u$ $u$ back into our result to get

$\int e^{tan x} {sec}^{2} (x) d x = e^{tan x} + C$ $int e^(tan x) sec^(2)(x) dx = e^(tan x) + C$

How do you integrate etanx⋅sec2(x)dxe^(tanx) * sec^2(x) dx?